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Math Help - Help of Intermediate Value Theorem

  1. #1
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    Help of Intermediate Value Theorem

    Use the Intermediate Value Theorem to show that x^5-x^2+1=0has a solution.

    Given:
    f(-1)=-1 and f(0)=1

    I understand what the IVT is and how to find a approximate solution using trial and error, but what exactly does it mean by "show?" Are they asking for a proof? I understand the concept of IVT but I can't translate my potential way of solving this into math symbols and such

    Thanks
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  2. #2
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    Quote Originally Posted by c_323_h
    Use the Intermediate Value Theorem to show that x^5-x^2+1=0has a solution.

    Given:
    f(-1)=-1 and f(0)=1
    The function,
    f(x)=x^5-x^2+1 is countinous on [-1,0]
    By IVT,
    For any  -1=f(-1)\leq k\leq f(0)=1 there exists an 1 \leq x \leq 0 such as, f(x)=k
    Let k=0 and the proof is finished.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    The function,
    f(x)=x^5-x^2+1 is countinous on [-1,0]
    By IVT,
    For any  -1=f(-1)\leq k\leq f(0)=1 there exists an 1 \leq x \leq 0 such as, f(x)=k
    Let k=0 and the proof is finished.
    This is what I tried. Let f(x)=x^5-x^2+1=0 . Then, by the IVT, there exists a number c such that
    f(-1)<f(c)<f(0). Since the function is continuous over the interval (-1,1) there exists a number N such that -1<N<1 and f(c)=N

    I don't know if that proves anything though. I really don't have a direction when I try to prove things. I just copied the definition of IVT out of the book.
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  4. #4
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    Let me try this again.

    If f(x) is a real function countinous on closed interval [a,b], and let k be any number between f(a) and f(b) then there must exist an x on [a,b] such as, f(x)=k.

    Root location theorem: If f(x) is countinous on [a,b] and f(a)f(b)<0 (in other words opposite signs) then there is a solution to the equation f(x)=0 on [a,b]

    Proof: By the condition we have that f(a) and f(b) have opposite signs. Thus, zero is a number between them. Thus, by the intermediate value theorem, f(x)=0 for some x on [a,b]
    ---
    Returning to your problem.
    1)Given f(x)=x^5-x^2+x+1 on [-1,0]
    2)The function is countinous cuz all polynomials are.
    3)And, f(-1)=-1 and f(0)=1
    4)And, f(-1)f(0)<0
    5)Condition for Root location theorem are satisfied.
    6)Thus there exists a solution to f(x)=0.
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